Chapter E, Problem 1E

To determine

To Calculate: ${\displaystyle \int x{e}^{-3x}dx}$

Answer to Problem 1E

  ${\displaystyle \int x{e}^{-3x}dx}=-\frac{e^{-3x}(3x+1)}{9}+C$ where C is constant

Explanation of Solution

Given:

  ${\displaystyle \int x{e}^{-3x}dx}$

Formula used:

The Integration by parts formula ${\displaystyle \int udv}=uv-{\displaystyle \int vdu}$ .

  ${\displaystyle \int e^{ax}dx}=\frac{1}{a}{e}^{ax}+C$ , where C is constant.

Calculation:

  ${\displaystyle \int x{e}^{-3x}dx}$ can be considered as ${\displaystyle \int x(e^{-3x}dx)}={\displaystyle \int udv}$

Where

  $u=x$ and $dv=e^{-3x}dx$

In this case differentiating $u$ gives:

  $du=dx$

and integrating $dv$ gives

  $v=\frac{e^{-3x}}{-3}$

Taking $C=0$ to obtain simplest antiderivative $\left(\because {\displaystyle \int e^{-3x}dx}=\frac{e^{-3x}}{-3}+C\right)$

Then by Integration by parts formula

  $\begin{array}{l}{\displaystyle \int (x)(e^{-3x}dx)}=-\frac{x{e}^{-3x}}{3}-{\displaystyle \int (-\frac{e^{-3x}}{3})}dx\left(\because {\displaystyle \int udv}=uv-{\displaystyle \int vdu}\right)\\ =-\frac{x{e}^{-3x}}{3}-\frac{e^{-3x}}{9}+C\text{ where }C\text{ is constant}\text{.}\end{array}$

After simplification

  ${\displaystyle \int (x)(e^{-3x}dx)}=-\frac{e^{-3x}(3x+1)}{9}+C\text{ where }C\text{ is constant}$